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#131




Peter O. Brackett wrote:
In a nutshell, photons do not cancel. :) Is this quote from _Optics_, by Hecht incorrect? "Unlike ordinary objects, photons cannot be seen directly; what is known of them comes from observing the results of their being either created or annihilated." Waves are an illusion, they are not the way Mother Nature works! Note that waves can pass through two slits at the same time. :) A footnote from _Optics_: "... although the lightiscorpuscular model has wide acceptance ... and is a crucial part of the contemporary discourse, like almost everything else in physics, it is not yet established beyond all doubt. See, for example, the summary article by R. Kidd, J. Ardini, and A. Anton, "Evolution of the modern photon", Am. J. Phys. 57 (1), 27 (1989)." Deal with it! Accept on faith the existence of God and virtual photons? :) To me, God and virtual photons seem to both be band aids for things not yet understood by man. To me, the following two statements don't seem all that different. 1. Although completely undetectable, virtual photons have to be the vehicle by which electrons communicate their presence to each other. 2. Although completely undetectable, God has to be the vehicle by which everything was created. It just occurred to me that virtual photons might explain telepathy. What do you think?  73, Cecil http://www.qsl.net/w5dxp = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 100,000 Newsgroups  19 Different Servers! = 
#132




2. Although completely undetectable, God has to be the vehicle by which everything was created. 73, Cecil http://www.qsl.net/w5dxp Last time when I tried to explain how the transistor radio works to my dog, she wiggled her tail: "It can't be and it don't taste no good" :) Yuri da BUm 
#133




Good comments.
One of my professors made an astute observation that sums up your conclusion. He said that science isn't set back so much by getting the wrong results by using the right methods as it is by getting the right results using the wrong methods. Amen to that. Roy Lewallen, W7EL David Robbins wrote: yes, i still hold that they are different. they are from completely different realms of electromagnetics. in the transmission line reflection coefficient you are working with a distributed system that is modeled with wave equations. there are delays, waves travel and reflections return after a finite delay. in your example you have coerced the voltage and current waves to be the same in the transmission line, but that is not the same as modeling the line itself. in the generator feeding a load you are looking at a lumped model. these are normally not solved with wave equations, there are no delays or reflections. the problem of calculating maximum power transfer from a generator to a load of a given impedance is a simple problem solved with no waves and only one pde used to find the max of the power to the load wrt the generator impedance. it doesn't care what the load is, as long as it is a linear time invariant impedance. now, you could create that impedance by putting a load on the end of a transmission line and transforming it back through the normal transmission line equations, but once you do that you throw the line out and solve the simple equations in the lumped models. these discussions always seem to end up in this same quagmire, one group trying to solve everything with wave equations, sinusoidal steady states, transmission line transformations, and the other holding on to the lumped models and trying to make them fit the wrong problem domain. one thing i did learn in ee classes was to use the right model for the job, choosing the wrong one will give you wrong answers most of the time.... the real problem is that sometimes you can get the right answer from the wrong model just because the numbers work out... and when you have to show your work you lose credit for using the wrong method but still getting the right numerical answer. that is what i see here, someone has started with a simplified case that just happens to work out to the same answer and has tried to generalize it... 10 points for the wrong method! 
#134




Dave:
[snip] these discussions always seem to end up in this same quagmire, one group trying to solve everything with wave equations, sinusoidal steady states, transmission line transformations, and the other holding on to the lumped models and trying to make them fit the wrong problem domain. one thing i did learn in ee classes was to use the right model for the job, choosing the wrong one will give you wrong answers most of the time.... the real problem is that sometimes you can get the right answer from the wrong model just because the numbers work out... and when you have to show your work you lose credit for using the wrong method but still getting the right numerical answer. that is what i see here, someone has started with a simplified case that just happens to work out to the same answer and has tried to generalize it... 10 points for the wrong method! [snip] Dave: I disagree, I feel that for measurements and calculations involving the impedance matching dynamics at any reference plane in such systems the differences between lumped and distributed systems is trivial and unimportant. The impedance matching dynamics at the reference plane/junction point are the same for distributed and lumped systems, they obey all the same equations, their electrodyanamics is the same, one simply cannot tell the difference at the driving point. If you don't like lumped models, then make the internal resistance of the Thevenin generator out of a distributed line. It will work just the same way. The disagreements that occur when there are discussions/arguments occur because folks often assume they are different situations, but they are not. Every line has a driving point impedance of Zo and at that point the voltage and current are related by Ohms Law v = Zo*i just as for a lumped system of impedance Zo. It simply doesn't matter if Zo is lumped or distributed when making observations at the driving point. And this is what one is doing when making impedance matching calculations. I agree that if you move off down the line in distance/space that you will find differences, but you certainly can't observe them at the driving points. A slotted line device for measuring VSWR is a physical example of a measurement system that works on a distributed system that won't work for a lumped system. However you can exactly correlate the measurements taken with the slotted line to measurements taken at the driving point. And so, although I agree with you that some can't seem to be able to see that these situations are the same, I don't agree that makes them different. They are the same at the driving point.  Peter K1PO Indialantic BytheSea, FL. 
#135




David Robbins wrote:
"Peter O. Brackett" wrote in message link.net... yes, i still hold that they are different. they are from completely different realms of electromagnetics. in the transmission line reflection coefficient you are working with a distributed system that is modeled with wave equations. there are delays, waves travel and reflections return after a finite delay. in your example you have coerced the voltage and current waves to be the same in the transmission line, but that is not the same as modeling the line itself. In simulation programs, transmission lines are solved for their twoport parameters, and are then treated as lumped circuits in the actual simulation, just like any lumpedelement circuit. Which is a good way to do it. I notice that in the ARRL Antenna Book, 19th edition , on page 247, it is stated with definite finality that the reflection coefficient formula uses the complex conjugate of Zo in the numerator. I also understand that this has been established by a "welltrusted authority". I have used Mathcad to calculate rho and VSWR for Reg's example, for many values of X0 (imaginary part of Z0) from 0 to 250 ohms. The data follows: Note: rho1* is conjugated rho1, SWR1 is for rho1*, rho2 is not conjugated and SWR2 applies to rho2 X0.......rho1*..SWR1.....rho2..SWR2 250..... 0.935...30.0.....1.865...3.30 200..... 0.937...30.8.....1.705...3.80 150..... 0.942...33.3.....1.517...4.87 100..... 0.948...37.5.....1.320...7.25 050..... 0.955...43.3.....1.131...16.3 020..... 0.959...47.6.....1.030...76.5 015..... 0.960...48.4.....1.010...204 012..... 0.960...48.9.....0.997....+/ infinity 010..... 0.960...49.2.....0.990....+305 004..... 0.961...76.3.....0.974....+76.3 0000..... 0.961...50.9.....0.961....+50.9 The numbers for notconjugate rho are all over the place and lead to ridiculous numbers for SWR. It is also obvious that for a lowloss line it doesn't matter much. But values of rho greater than 1.0, on a Smith chart correspond to negative values of resistance (see the data). Something is wrong here that we are overlooking. The use of conjugate rho is so much better behaved that I have some real doubts about some of our conclusions on this matter. What about it folks? How can we get to the bottom of this? Bill W0IYH 
#136




From your more recent posting, I misinterpreted what you mean by
"driving point impedance". It's simply the impedance of the transmission line, not the impedance from the load looking back toward the source, nor the source impedance. Based on that interpretation, I'll try again. Peter O. Brackett wrote: . . . Now at any particular frequency w = 2*p*f you will find that this general complex surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx. Later let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical example. Consider either a semiinfinite length of this Zo line, or even a finite length of the Zo line terminated in an impedance equal to Z0. I am sure that you will agree that both the semiinfinite Zo line or the finite length Zo line terminated in Zo have the same driving point impedance namely Zo. That is the impedance of the line. Ok. Now excite this semiinfinte Zo line by an ideal generator of open circuit voltage Vi = 2*a behind an impedance equal to the surge impedance Zo. In other words this is a Thevenin generator of ideal constant voltage 2a behind a complex impedance of Zo. Oops. No it's not. Consider circuit "A", a voltage source in series with a lumped impedance equal to r + jx. And circuit "B", a voltage source in series with a transmission line of length, say, one wavelength, with characteristic impedance Z0 = r + jx. The two are not at all equivalent. To show that they're not, connect a 50 ohm resistor to the output of each. The resistor current in circuit "A" is Vi/(50 + r + jx). The current in circuit "B" is Vi/50. David Robbins recently made several postings explaining that these two circuits are not equivalent. Hopefully the above example illustrates that. Can't you do your "proof" with, say, a one wavelength transmission line of characteristic impedance Z0? Roy Lewallen, W7EL 
#137




William E. Sabin wrote:
David Robbins wrote: "Peter O. Brackett" wrote in message link.net... yes, i still hold that they are different. they are from completely different realms of electromagnetics. in the transmission line reflection coefficient you are working with a distributed system that is modeled with wave equations. there are delays, waves travel and reflections return after a finite delay. in your example you have coerced the voltage and current waves to be the same in the transmission line, but that is not the same as modeling the line itself. In simulation programs, transmission lines are solved for their twoport parameters, and are then treated as lumped circuits in the actual simulation, just like any lumpedelement circuit. Which is a good way to do it. I notice that in the ARRL Antenna Book, 19th edition , on page 247, it is stated with definite finality that the reflection coefficient formula uses the complex conjugate of Zo in the numerator. I also understand that this has been established by a "welltrusted authority". I have used Mathcad to calculate rho and VSWR for Reg's example, for many values of X0 (imaginary part of Z0) from 0 to 250 ohms. The data follows: Note: rho1* is conjugated rho1, SWR1 is for rho1*, rho2 is not conjugated and SWR2 applies to rho2 X0.......rho1*..SWR1.....rho2..SWR2 250..... 0.935...30.0.....1.865...3.30 200..... 0.937...30.8.....1.705...3.80 150..... 0.942...33.3.....1.517...4.87 100..... 0.948...37.5.....1.320...7.25 050..... 0.955...43.3.....1.131...16.3 020..... 0.959...47.6.....1.030...76.5 015..... 0.960...48.4.....1.010...204 012..... 0.960...48.9.....0.997....+/ infinity 010..... 0.960...49.2.....0.990....+305 004..... 0.961...76.3.....0.974....+76.3 0000..... 0.961...50.9.....0.961....+50.9 The numbers for notconjugate rho are all over the place and lead to ridiculous numbers for SWR. It is also obvious that for a lowloss line it doesn't matter much. But values of rho greater than 1.0, on a Smith chart correspond to negative values of resistance (see the data). Something is wrong here that we are overlooking. The use of conjugate rho is so much better behaved that I have some real doubts about some of our conclusions on this matter. What about it folks? How can we get to the bottom of this? Bill W0IYH The equation in the ARRL Antenna Book is identical to the equation for rho that is in the Power Wave literature (see Gonzalez and also see Kurokawa). Also, numerous literature sources describe how an opencircuit generator with internal impedance Z0, connected directly to load ZL, is actually a power wave setup that leads to a rho formula that is identical to the formula in the ARRL Antenna Book. When calculating rho, it is not necessary to fool around with the wave equations, because frequency is constant and everything is steadystate. Bill W0IYH 
#138




Roy:
[snip] Oops. No it's not. Consider circuit "A", a voltage source in series with a lumped impedance equal to r + jx. And circuit "B", a voltage source in series with a transmission line of length, say, one wavelength, with characteristic impedance Z0 = r + jx. The two are not at all equivalent. To show that they're not, connect a 50 ohm resistor to the output of each. The resistor current in circuit "A" is Vi/(50 + r + jx). The current in circuit "B" is Vi/50. [snip] Y'all are missing my point. First off... you, and Dave have just changed situation I proposed. I did not say that there was an unterminated line of one wavelength! You and Dave said that. Forget it. That has nothing to do with the discussion/proof! Read my "typing"! I said you must have *either* a semiinfinite line, whose driving point impedance is known by all to be exactly Zo, *or*, but it is simply beside the point, if you must insist on using a finite length line then that is OK as well as long as it is terminated at it's far end in an impedance equal to Zo. In either case of the semiinfinite Zo line or the finite line terminated in Zo the driving point impedance is identical, i.e. it is Zo and you can't tell the difference between the two at the driving point. [snip] David Robbins recently made several postings explaining that these two circuits are not equivalent. Hopefully the above example illustrates that. [snip] I disagree with Dave. A semiinfinite Zo line and a finite Zo line terminated in Zo are indistinguisable at the driving point and at all points along the lines, until you come to the end of the finite line, but even there the voltage and current across then terminating Zo is identical to the voltage and current at the same point on the semiinfinite line. All of this discussion about there being a difference is moot. There is no difference. That's exactly why I set it up that way, simply because there is no difference. You guys are missing the point. Check out any book on transmission lines... When you look into the end of a transmission line of semiinfinite extent you see a driving point impedance of Zo. That is the very definition of Zo! If you cut the line off to a finite length and terminate it in it's characteristic impedance Zo the driving point remains the same, i.e. Zo. That again is part of the defining charateristic of characteristic or surge impedance! Take a chunk of 50 Ohm lossless coax and terminate it in 50 Ohms. What's the impedance you see looking into the end? 50 Ohms. Now change the length to something else and again terminate it in 50 Ohms, what 's the impedance? 50 Ohms, it doesn't matter how long the line is if it is terminated in it's characteristic impedance it's driving point impedance is Zo! End of story. You simply can't tell the difference between a line terminated in it's characteristic impedance and the characteristic impedance of an semiinfinite line as found by observiations at the driving point, they are one and the same. Zo! [snip] Can't you do your "proof" with, say, a one wavelength transmission line of characteristic impedance Z0? Roy Lewallen, W7EL [snip] Yep, sure can... sigh, why is this soooo hard. Use a one wavelength line if you insist, but terminate it at the far end with an impedance equal to Zo!, The length of the line has nothing to do with the proof. As a matter of fact that is one of the neat things about the socalled "proof" I cooked up. The "proof" is just about whether there are reflections at a conjugate match or not. The proof, mathematical or experimental, just proves that there are no reflections [as defined by the classical rho] when the generator sees it's image Zo, but there are reflections when the generator sees it's conjugate! And the same in the reverse direction. i.e. there are no reflections when the line sees it's image Zo but there are reflections when it sees it's conjugate. And if Zo is a real resistance say Zo = R, then both situations are identical. That's all, nothing complicated, simple straightforward stuff, when Zo is complex, there *will* always be reflections at a conjugate match. This means that when there is a conjugate match on a complex line a "classical" reflectometer will not indiacate zero reflected voltage. And if the scale on the reflectometer is actually calibrated in "Watts" even though it measures Volts, it will indicate the presence of [[a "false"] reflected power at a conjugate match. But then everybody already knew that! QED!  Peter K1PO Indialantic BytheSea, FL. 
#139




Not QED at all. You claimed to have proved that maximum power is
delivered to a load when a transmission line is terminated such that the reflected voltage on the line is zero. Here, you're agreeing that the reflected voltage is zero when the line is terminated in its characteristic impedance. So where's the proof that this condition leads to maximum power to the load? Roy Lewallen, W7EL Peter O. Brackett wrote: Roy: [snip] Oops. No it's not. Consider circuit "A", a voltage source in series with a lumped impedance equal to r + jx. And circuit "B", a voltage source in series with a transmission line of length, say, one wavelength, with characteristic impedance Z0 = r + jx. The two are not at all equivalent. To show that they're not, connect a 50 ohm resistor to the output of each. The resistor current in circuit "A" is Vi/(50 + r + jx). The current in circuit "B" is Vi/50. [snip] Y'all are missing my point. First off... you, and Dave have just changed situation I proposed. I did not say that there was an unterminated line of one wavelength! You and Dave said that. Forget it. That has nothing to do with the discussion/proof! Read my "typing"! I said you must have *either* a semiinfinite line, whose driving point impedance is known by all to be exactly Zo, *or*, but it is simply beside the point, if you must insist on using a finite length line then that is OK as well as long as it is terminated at it's far end in an impedance equal to Zo. In either case of the semiinfinite Zo line or the finite line terminated in Zo the driving point impedance is identical, i.e. it is Zo and you can't tell the difference between the two at the driving point. [snip] David Robbins recently made several postings explaining that these two circuits are not equivalent. Hopefully the above example illustrates that. [snip] I disagree with Dave. A semiinfinite Zo line and a finite Zo line terminated in Zo are indistinguisable at the driving point and at all points along the lines, until you come to the end of the finite line, but even there the voltage and current across then terminating Zo is identical to the voltage and current at the same point on the semiinfinite line. All of this discussion about there being a difference is moot. There is no difference. That's exactly why I set it up that way, simply because there is no difference. You guys are missing the point. Check out any book on transmission lines... When you look into the end of a transmission line of semiinfinite extent you see a driving point impedance of Zo. That is the very definition of Zo! If you cut the line off to a finite length and terminate it in it's characteristic impedance Zo the driving point remains the same, i.e. Zo. That again is part of the defining charateristic of characteristic or surge impedance! Take a chunk of 50 Ohm lossless coax and terminate it in 50 Ohms. What's the impedance you see looking into the end? 50 Ohms. Now change the length to something else and again terminate it in 50 Ohms, what 's the impedance? 50 Ohms, it doesn't matter how long the line is if it is terminated in it's characteristic impedance it's driving point impedance is Zo! End of story. You simply can't tell the difference between a line terminated in it's characteristic impedance and the characteristic impedance of an semiinfinite line as found by observiations at the driving point, they are one and the same. Zo! [snip] Can't you do your "proof" with, say, a one wavelength transmission line of characteristic impedance Z0? Roy Lewallen, W7EL [snip] Yep, sure can... sigh, why is this soooo hard. Use a one wavelength line if you insist, but terminate it at the far end with an impedance equal to Zo!, The length of the line has nothing to do with the proof. As a matter of fact that is one of the neat things about the socalled "proof" I cooked up. The "proof" is just about whether there are reflections at a conjugate match or not. The proof, mathematical or experimental, just proves that there are no reflections [as defined by the classical rho] when the generator sees it's image Zo, but there are reflections when the generator sees it's conjugate! And the same in the reverse direction. i.e. there are no reflections when the line sees it's image Zo but there are reflections when it sees it's conjugate. And if Zo is a real resistance say Zo = R, then both situations are identical. That's all, nothing complicated, simple straightforward stuff, when Zo is complex, there *will* always be reflections at a conjugate match. This means that when there is a conjugate match on a complex line a "classical" reflectometer will not indiacate zero reflected voltage. And if the scale on the reflectometer is actually calibrated in "Watts" even though it measures Volts, it will indicate the presence of [[a "false"] reflected power at a conjugate match. But then everybody already knew that! QED!  Peter K1PO Indialantic BytheSea, FL. 
#140




"Peter O. Brackett" wrote in message hlink.net... Roy: This is a long winded explanation of a simple concept comprising a very simple circuit containing an ideal generator, a complex impedance and a terminated line, along with no more than a couple of equations that constitutes a clear and simple mathematical and/or experimental "proof" that the reflected voltage is nonzero for a conjugate match and is zero for an image match. QED sorry, not always... another clear generalization from oversimplifying and applying the wrong model. there are cases where there is zero reflected voltage for a conjugate match. 
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